How many 6 digit numbers can formed by 0-5 digits divisible by digit at unit place? options1-420 2-426 3-432,none

• 1 at unit place=>4*4*3*2*1=96
  2 at unit place=>4*4*3*2*1=96
  3 at unit place=>4*4*3*2*1=96
  5 at unit place=>4*4*3*2*1=96 but for 4 the last 2 digit must be divisible by 4 so, we have 2 options 0 and 2 at second place and 4 at unit place. now,
  when 4 at unit place and 2 at second place=>3*3*2*1*1=18
  and 4 at unit place and 0 at second place=>4*3*2*1*1=24
  Now, 96*4+18+24=426
• 10 years agoHelpfull: Yes(19) No(7)
• _ _ _ _ _ 1 => one at unit place, all 6digit number are divisible by one,
  but 1 should be at once place and we can manipulate rest 5 digits in 5! ways
  therefore 5!=120
  _ _ _ _ _ 2 => 2 at units place, all number ending with 2 are divisible by 2,
  therefore 5!=120
  _ _ _ _ _ 3=> 1+2+4+5+6+3=21 divisible by 3
  therefore 5!=120
  _ _ _ _ (2/6) 4 => 4 at unit place and either 2 or 6 in tens place make the whole number divisible by 4, rest 4 digits in 4! ways
  therefore 4!*2=48
  _ _ _ _ _ 5 => 5!= 120
  _ _ _ _ _ 6 => 5!= 120
  
  total comes up to 120+120+120+120+120+48=648
• 10 years agoHelpfull: Yes(0) No(11)
• Ans is 426.
  focus on zero(As leftmost digit cannot be zero)
  a)- - - - - 1
  4 4 3 2 1 1 -> divisibility by 1
  total = 96
  b)- - - - - 2
  4 4 3 2 1 2 -> divisibility by 2
  Total = 96
  c)- - - - - 3
  4 4 3 2 1 3 -> divisibility by 3
  Total 96
  d)- - - - - 4
  Here, take 2 cases for divisibilty by 4.
  - - - - 2 4
  3 3 2 1 2 4
  Total 18
  Another case as,
  - - - - 2 0
  4 3 2 1 2 0
  Total = 24
  e) - - - - - 5
  4 4 3 2 1 5 -> divisibilty by 5
  Total = 96
• 7 years agoHelpfull: Yes(0) No(0)